Respuesta :
Better to see it for yourself :) .
[tex]3^{x-1} = 3^{4 - x}[/tex]
can be solved easily by taking the logarithm in base 3 on both sides. We can do that because the log and the exponential functions are strictly increasing. It gives
[tex]x - 1 = 4 - x[/tex]
which is easy to solve.
BUT if you have
[tex]3^{x - 1} = 4^{x - 2}[/tex]
then you cannot simply take the logarithm of one base. Let's see why.
[tex]log_{3}(3^{x - 1}) = log_{3}(4^{x - 1})[/tex]
gives
[tex]x - 1 = log_{3}(4^{x - 1})[/tex]
and the simplification is not easy. I think the easiest way (if not the only) is to change the base of the logarithm with the formula
[tex]log_{a}x = \frac{log_{b}x}{log_{b}a}[/tex]
In our example it gives
[tex]x - 1 = \frac{log_{4}(4^{x - 1})}{log_{4}3}[/tex]
Which yields
[tex]x - 1 = \frac{x - 1}{log_{4}3}[/tex]
From there, it is a simple degree one equation.
[tex]3^{x-1} = 3^{4 - x}[/tex]
can be solved easily by taking the logarithm in base 3 on both sides. We can do that because the log and the exponential functions are strictly increasing. It gives
[tex]x - 1 = 4 - x[/tex]
which is easy to solve.
BUT if you have
[tex]3^{x - 1} = 4^{x - 2}[/tex]
then you cannot simply take the logarithm of one base. Let's see why.
[tex]log_{3}(3^{x - 1}) = log_{3}(4^{x - 1})[/tex]
gives
[tex]x - 1 = log_{3}(4^{x - 1})[/tex]
and the simplification is not easy. I think the easiest way (if not the only) is to change the base of the logarithm with the formula
[tex]log_{a}x = \frac{log_{b}x}{log_{b}a}[/tex]
In our example it gives
[tex]x - 1 = \frac{log_{4}(4^{x - 1})}{log_{4}3}[/tex]
Which yields
[tex]x - 1 = \frac{x - 1}{log_{4}3}[/tex]
From there, it is a simple degree one equation.
Answer:
because it is less work to convert one of the sides to match the others because you can't do anywork without matching bases.
Step-by-step explanation:
